Toán Lớp 6: B2:tìm K ∈ Q,biết b,2(2/5-K)=4/7 c,3K-(-1/8)=6/5 d,K+1/2028+K+1/2019+K+1/2020+K+1/2021=0

Question

Toán Lớp 6:  B2:tìm K ∈ Q,biết  b,2(2/5-K)=4/7 c,3K-(-1/8)=6/5 d,K+1/2028+K+1/2019+K+1/2020+K+1/2021=0

cattien · Answer

Giải đáp + Lời giải và giải thích chi tiết: b) 2(2/5-k)=4/7 <=>2/5-k=4/7:2 <=>2/5-k=2/7 <=>k=2/5-2/7 <=>k=14/35-10/35=4/35 Vậy k = 4/35 c) 3k-(-1/8)=6/5 <=>3k+1/8=6/5 <=>3k=6/5-1/8 <=>3k=43/40 <=>k=43/40:3=43/120 Vậy k = 43/120 d) Trường hợp 1 : k+1/2028+k+1/2019+k+1/2020+k+1/2021=0 <=>4k+(1/2028+1/2019+1/2020+1/2021)=0 -> Không tìm được k vì số quá dữ Trường hợp 2 : (k+1)/2028+(k+1)/2019+(k+1)/2020+(k+1)/2021=0 <=>(k+1)(1/2028+1/2019+1/2020+1/2021)=0 <=>k+1=0<=>k=-1 Vì 1/2028+1/2019+1/2020+1/2021ne0 Vậy k = -1.

ngocle44 · Answer

$  $   b,   2 (2/5 - k) =4/7   -> 2/5 - k=4/7: 2   -> 2/5 - k = 2/7   ->k=2/5 - 2/7   -> k=4/35 (Thỏa m&atilde;n)   Vậy k=4/35   $  $   c,   3k - (-1)/8 = 6/5   ->3k=6/5 + (-1)/8   ->3k=43/40   ->k=43/40 : 3   ->k=43/120 (Thỏa m&atilde;n)   Vậy k=43/120   $  $   d,   (k+1)/2028 + (k+1)/2019 + (k+1)/2020 + (k+1)/2021=0   -> (k+1) (1/2028 + 1/2019 + 1/2020 + 1/2021)=0   ->k+1=0 (Do 1/2028 + 1/2019 + 1/2020 + 1/2021  ne 0)   ->k=0-1   ->k=-1 (Thỏa m&atilde;n)   Vậy k=-1